问题描述
为什么通过void*删除对象是未定义的行为，而不是编译错误?
Why would deleting an object through void* be undefined behavior, rather than compilation error?
void foo(void* p) {
delete p;
}
此代码编译并生成代码，尽管带有gcc和clang的警告(令人惊讶的是，ICC没有给出警告):
This code compiles and produces code, albeit with the warning on gcc and clang (surprisingly, ICC doesn't give a warning):
为什么不是简单的格式无效的语法无效的程序?看起来Standard并没有花太多时间在[expr.delete]上
Why is not simply malformed program with invalid syntax? Looks like Standard doesn't spend too much time on it, saying in [expr.delete] that
我会因为什么原因而不会触发硬编译错误吗?
Would there be any reason I am missing why this does not trigger a hard compilation error?
推荐答案
在现代C ++中，删除void *指针 的格式不正确(即我们通常所说的编译错误" )
In modern C++ deleting a void * pointer is ill-formed (i.e. it is what we typically call a "compilation error")
void *不是对象类型的指针.
void * is not a pointer to object type.
在C ++ 98中，情况有所不同.删除void *类型的空指针是NOP，而删除void *类型的非空指针则是UB.
In C++98 the situation was different. Deleting a null pointer of void * type was a NOP, while deleting a non-null pointer of void * type was UB.
此规范更改似乎是由缺陷报告#599 .原始规范允许在delete-expression中提供 any 指针类型的空指针，例如函数指针.这看起来是不必要的. DR#599的分辨率提高了要求，同时也禁止了void *.
This change in the specification appears to have been triggered by defect report #599. The original spec allowed supplying null pointers of any pointer type in delete-expression, like function pointers, for example. This looked unnecessarily permissive. The resolution of DR#599 tightened the requirements, outlawing void * as well.
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